OCR A Level Physics

Revision Notes

5.5.5 Acceleration & Displacement

Test Yourself

Acceleration & Displacement of an Oscillator

  • The acceleration of an object oscillating in simple harmonic motion is:

a = −⍵2x

  • Where:
    • a = acceleration (m s-2)
    • ⍵ = angular frequency (rad s-1)
    • x = displacement (m)

  • This is used to find the acceleration of an object in SHM with a particular angular frequency ⍵ at a specific displacement x
  • The equation demonstrates:
    • The acceleration reaches its maximum value when the displacement is at a maximum ie. x = x0 (amplitude)
    • The minus sign shows that when the object is displacement to the right, the direction of the acceleration is to the left

Graph of acceleration and displacement, downloadable AS & A Level Physics revision notes

The acceleration of an object in SHM is directly proportional to the negative displacement

  • The graph of acceleration against displacement is a straight line through the origin sloping downwards (similar to y = − x)
  • Key features of the graph:
    • The gradient is equal to − ⍵2
    • The maximum and minimum displacement x values are the amplitudes −x0 and +x0

  • A solution to the SHM acceleration equation is the displacement equation:

x = x0sin(⍵t)

  • Where:
    • x = displacement (m)
    • x0 = amplitude (m)
    • t = time (s)

  • This equation can be used to find the position of an object in SHM with a particular angular frequency and amplitude at a moment in time
    • Note: This version of the equation is only relevant when an object begins oscillating from the equilibrium position (x = 0 at t = 0)

  • The displacement will be at its maximum when sin(⍵t) equals 1 or − 1, when x = x0
  • If an object is oscillating from its amplitude position (x = x0 or x = − x0 at t = 0) then the displacement equation will be:

x = x0cos(⍵t)

  • This is because the cosine graph starts at a maximum, whilst the sine graph starts at 0

Displacement SHM graphs, downloadable AS & A Level Physics revision notes

These two graphs represent the same SHM. The difference is the starting position

Worked example

A mass of 55 g is suspended from a fixed point by means of a spring. The stationary mass is pulled vertically downwards through a distance of 4.3 cm and then released at t = 0.

The mass is observed to perform simple harmonic motion with a period of 0.8 s.

Calculate the displacement x in cm of the mass at time t = 0.3 s.

Step 1: Write down the SHM displacement equation

    • Since the mass is released at t = 0 at its maximum displacement, the displacement equation will be with the cosine function:

x = x0 cos(⍵t)

Step 2: Calculate angular frequency

straight omega space equals space fraction numerator 2 straight pi over denominator straight T end fraction space equals space fraction numerator 2 straight pi over denominator 0.8 end fraction space equals space 7.85 space rad space straight s to the power of negative 1 end exponent

    • Remember to use the value of the time period given, not the time where you are calculating the displacement from

Step 3: Substitute values into the displacement equation

x = 4.3 cos (7.85 × 0.3) = –3.0369… = –3.0 cm (2 s.f)

    • Make sure the calculator is in radians mode
    • The negative value means the mass is 3.0 cm on the opposite side of the equilibrium position to where it started (3.0 cm above it)

Exam Tip

Since displacement is a vector quantity, remember to keep the minus sign in your solutions if they are negative, you could lose a mark if not!

Also, remember that your calculator must be in radians mode when using the cosine and sine functions. This is because the angular frequency ⍵ is calculated in rad s-1, not degrees.

You often have to convert between time period T, frequency f and angular frequency ⍵ for many exam questions – so make sure you revise the equations relating to these.

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